Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/2278.html

2278. Percentage of Letter in String

• Difficulty: Easy.
• Related Topics: String.
• Similar Questions: Sort Characters By Frequency.

Problem

Given a string s and a character letter, return** the percentage of characters in s that equal letter rounded down to the nearest whole percent.**

  Example 1:

Input: s = "foobar", letter = "o"
Output: 33
Explanation:
The percentage of characters in s that equal the letter 'o' is 2 / 6 * 100% = 33% when rounded down, so we return 33.

Example 2:

Input: s = "jjjj", letter = "k"
Output: 0
Explanation:
The percentage of characters in s that equal the letter 'k' is 0%, so we return 0.

  Constraints:

• 1 <= s.length <= 100

• s consists of lowercase English letters.

• letter is a lowercase English letter.

Solution (Java, C++, Python)

• Java
• Python
• C++
• Go
• TypeScript
• RenderScript

• class Solution {
      public int percentageLetter(String s, char letter) {
          int count = 0;
          int n = s.length();
          for (int i = 0; i < n; i++) {
              if (s.charAt(i) == letter) {
                  ++count;
              }
          }
          return (count * 100) / (n);
      }
  }
  
  ############
  
  class Solution {
      public int percentageLetter(String s, char letter) {
          int cnt = 0;
          for (char c : s.toCharArray()) {
              if (c == letter) {
                  ++cnt;
              }
          }
          return cnt * 100 / s.length();
      }
  }
  

• class Solution:
      def percentageLetter(self, s: str, letter: str) -> int:
          return s.count(letter) * 100 // len(s)
  
  ############
  
  # 2278. Percentage of Letter in String
  # https://leetcode.com/problems/percentage-of-letter-in-string/
  
  class Solution:
      def percentageLetter(self, s: str, letter: str) -> int:
          n = len(s)
          c = s.count(letter)
  
          return int(c / n * 100)
  
  

• class Solution {
  public:
      int percentageLetter(string s, char letter) {
          int cnt = 0;
          for (char& c : s) cnt += c == letter;
          return cnt * 100 / s.size();
      }
  };
  

• func percentageLetter(s string, letter byte) int {
  	cnt := 0
  	for i := range s {
  		if s[i] == letter {
  			cnt++
  		}
  	}
  	return cnt * 100 / len(s)
  }
  

• function percentageLetter(s: string, letter: string): number {
      let count = 0;
      let total = s.length;
      for (let i of s) {
          if (i === letter) count++;
      }
      return Math.floor((count / total) * 100);
  }
  
  

• impl Solution {
      pub fn percentage_letter(s: String, letter: char) -> i32 {
          let mut count = 0;
          for c in s.chars() {
              if c == letter {
                  count += 1;
              }
          }
          (count * 100 / s.len()) as i32
      }
  }
  
  

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).

All Problems

All Solutions